A Note About the Base of a Product Topology A question came up in class about the base of a product topology. This document hopefully addresses it after first establishing a few things... Consider two topological spaces (X; TX ) and (Y; TY ) and the product space Z = X × Y: A collection of sets BX ⊆ TX is a base (basis) for TX if every G 2 TX can be written as a union of sets in BX . For example, [ G = Bα α2I where I is some arbitrary, and possibly uncountable, index set, and fBαgα2I is a collection of sets in BX . The product space Z can be endowed with the product topology which we will denote here by TZ . It is defined as the topology whose base is the collection of sets fGX × GY : GX 2 TX ;GY 2 TY g: Clearly there are other possible topologies for Z. For example, we could take the trivial topology f;; Zg. However, the product topology will be an important one for us! The question was: Can we just take the base for TZ to be sets of the form fBX × BY : BX 2 BX ;BY 2 BY g? The answer is yes and we will prove it here shortly. Forget product spaces for a few moments. Suppose that (X; TX ) is a topological space. It will be useful to use an alternate definition of a base for T . For clarity, we restate our first definition and label it as “Definition 1". Definition 1: A collection of sets B ⊆ T is a base for T if every G 2 T can be written as a union of sets in B. Definition 2: A collection of sets B ⊆ T is a base for T if for any G 2 T and for any x 2 G, there exists a B 2 B such that x 2 B ⊆ G. Proof that Def 1 ) Def 2: • Take any G 2 T and any x 2 G. S • By Definition 1, we can write G = α2I Bα for some sets Bα 2 B and some index set I. • So, [ x 2 G = Bα ) x 2 Bα∗ α2I ∗ for some Bα∗ 2 B with α 2 I. [ • But Bα∗ ⊆ Bα = G, so we have α2I x 2 Bα∗ ⊆ G for some Bα∗ 2 B. ut Proof that Def 2 ) Def 1: • Take any G 2 T and any x 2 G. • By Definition 2, there exists a B 2 B such that x 2 B ⊆ G. • Denote B by Bx and find such base sets for all x 2 G. • Then clearly [ G ⊆ Bx x2G since every x 2 G is contained in Bx. • On the other hand, [ Bx ⊆ G x2G since each Bx ⊆ G. [ [ • Thus, we have G = Bx, so we have written G in the form Bα. ut x2G α We are ready to prove the following Theorem. Theorem: Let (X; TX ) and (Y; TY ) be two topological spaces with respective bases BX and BY . Let Z = X × Y. Then BX × BY := fBX × BY : BX 2 BX ;BY 2 BY g? is a base for the product topology TZ . Proof: Using Definition 2 of a topology base, We will show that, for any G 2 TZ and for any z = (x; y) 2 G, we can find BX 2 BX and BY 2 BY such that z 2 BX × BY ⊆ G. • Take any G 2 TZ and any z = (x; y) 2 G. • By Definition 2 of the product topology, there exists a GX × GY with GX 2 TX and GY 2 TY such that (x; y) 2 GX × GY ⊆ G: • Since BX and BY are respective bases for TX and TY , there exist sets C 2 BX and D 2 BY such that x 2 C ⊆ BX and y 2 D ⊆ BY : • Since we found C × D in BX × BY such that z = (x; y) 2 C × D ⊆ G, we have, by Definition 2, that BX × BY is a base for the product topology TZ . ut.